Estimation and Limits -- Adding Functions

You should use one of the computer algebra systems below with this module. Click on the appropriate icon for your preferred CAS and then arrange your screen so that you can easily move back-and-forth between this window and your CAS window. Click on the appropriate help button for help.

Adding two function has many practical implications. For example, if f(x) represents one television image and g(x) represents another television image then f(x) + g(x) represents the two signals superimposed on each other. We can combine this operation with the operation of multiplying a function by a constant to achieve some interesting effects. Continuing the example above, the sequence of functions

f(x)
0.9 f(x) + 0.1 g(x)
0.8 f(x) + 0.2 g(x)
         .
         .
0.1 f(x) + 0.9 g(x)
g(x)

represents a sequence of television images starting with f(x) and then making a smooth transition to g(x). This is frequently used in television and movies to make a transition from one scene to another.

f(x) = sin x
g(x) = x
h(x) = x + sin x

Use your CAS window to compare the following functions

f(x) = sin x
g(x) = -x
h(x) = -x + sin x

Use your CAS window to compare the following functions

f(x) = sin x
g(x) = x/2
h(x) = x/2 + sin x

Use your CAS window to compare the following functions

f(x) = 1/2^x
g(x) = 1
h(x) = 1 + 1/2^x

The basic fact that we discuss here is that if

  Lim   f(x) = L
x --> a

and

  Lim   g(x) = M
x --> a

then

  Lim   f(x) + g(x) = L + M
x --> a

We can express this same fact by saying that if f(x) is close to L when x is close to a and if g(x) is close to M when x is close to a then f(x) + g(x) is close to L + M when x is close to a.

However we express it, this descriptive fact is only part of a more interesting story. The first hypothesis -- the fact that

  Lim   f(x) = L
x --> a

requires that we can make f(x) as close as we want to L by making x sufficently close to a. The second hypothesis -- the fact that

  Lim   g(x) = M
x --> a

requires that we can make g(x) as close as we want to M by making x sufficiently close to a. The conclusion -- the fact that

  Lim   f(x) + g(x) = L + M
x --> a

tells us that we can make f(x) + g(x) as close as we want to L + M by making x sufficiently close to a. The proof of this fact is based on the observation that adding two functions not only adds their values at a but it also adds any errors. See the figure below.

Suppose that we are given a permissible error epsilon for f(x) + g(x). Because we want the total error to be less than epsilon, we will arrange things so that each of f(x) and g(x) has an error of no more than epsilon/2. This is easily accomplished.

Since

  Lim   f(x) = L
x --> a

there is a delta1 such that if

|x - a| < delta1

then

|f(x) - L| < epsilon/2

and since

  Lim   g(x) = M
x --> a

there is a delta2 such that if

|x - a| < delta2

then

|g(x) - M| < epsilon/2

Now let delta be the smaller of delta1 and delta2. Now if

|x - a| < delta

then

|f(x) - L| < epsilon/2

and

|g(x) - M| < epsilon/2

so

|f(x) + g(x) - (L + M)| =  |f(x) - L + g(x) - M| 
                        <= |f(x) - L| + |g(x) - M|
                        <  epsilon/2 + epsilon/2 = epsilon

Check Your Understanding

In manufacturing there is often a trade-off between precision and cost. The table below, for example, gives the cost of manufacturing a particular automobile part to a given tolerance -- that is, with a given permissible error.

 Tolerance     Cost

0.10 inches   $2.00
0.05 inches   $6.00
0.02 inches  $25.00
0.01 inches  $85.00

1. Suppose that two of these parts are connected end-to-end and that the tolerance for the total length is 0.10 inches. How much will it cost to manufacture the two parts?

   answer

2. Suppose that two of these parts are connected end-to-end and that the tolerance for the total length is 0.02 inches. How much will it cost to manufacture the two parts?

   answer